$\begingroup$Do you have some conjecture, counterexample? something?$\endgroup$
– sinbadhFeb 4 '16 at 0:13

$\begingroup$I'm inheritantly trying to show that for a normed space $\mathbb{X}$ a map from it's dual space (a complete space) to the space of continuously differentiable functions on $[0,1]$ (not a complete space) cannot be bijective$\endgroup$
– Bruno SternerFeb 4 '16 at 0:16

2 Answers
2

Edit: In fact there's a very simple theorem here that gives the whole truth: Given a bounded linear bijection $T:X\to Y$, where $X$ is complete, $Y$ is complete if and only if $T^{-1}$ is bounded. (If $Y$ is complete the open mapping theorem shows that $T^{-1}$ is bounded. On the other hand if $T^{-1}$ is bounded it's trivial to show that $Y$ is complete: A Cauchy sequence in $Y$ comes from a Cauchy sequence in $X$, which converges...)

Original:

Yes, it's possible. This surprises me; I thought the answer was no. The reason I thought the answer was no was something like this:

Let's agree that an isomorphism in the present context is a bounded linear bijection whose inverse is also bounded. Now (i) a bounded linear bijection between Banach spaces must be an isomorphism, (ii) if $X$ and $Y$ are isomorphic normed spaces and $X$ is complete then $Y$ is complete. Of course I never thought that was actually a proof here; all it proves is that $Y$ is complete if $Y$ is complete. But those facts in my head made me think the answer was no.

Then $T$ is certainly bounded and injective. Now let $Y=T(X)$, and give $Y$ the norm it inherits from $\ell^2$. Regard $T$ as a map from $X$ to $Y$. It's still bounded and injective, and now it's surjective.

So $T:X\to Y$ is a bounded linear bijection. And $Y$ is not complete. (Proof: If $Y$ were complete then the open mapping theorem would show that $T^{-1}:Y\to X$ was bounded, but $T^{-1}$ is certainly not bounded.)

$\begingroup$I believe that one. Really the same as my example - if you have any bounded linear injective $T:X\to Z$ which is not bounded below you get an example by setting $Y=T(X)$.$\endgroup$
– David C. UllrichFeb 4 '16 at 0:46

$\begingroup$Yes, the trick is finding a proper dense subspace which is Banach w.r.t. a different norm, and such that the inclusion is continuous. another way would be with the inclusions $L^p(0,1)\subsetneq L^q(0,1)$ for $q<p$. I was finishing writing this when you posted yours, and I deemed it worth mentioning because the maps involved are very well-known.$\endgroup$
– user228113Feb 4 '16 at 0:54

$\begingroup$Certainly worth mentioning - happens all the time that one post appears while another is being typed....$\endgroup$
– David C. UllrichFeb 4 '16 at 1:03

$\begingroup$Nice example.The larger norm is complete.Since it is larger, its topology is stronger so the identity embedding to the weaker-normed space is bounded (and continuous) and the weaker norm is incomplete, as it sits inside a Banach space as a dense proper subset.$\endgroup$
– DanielWainfleetFeb 4 '16 at 10:04